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mixer [17]
3 years ago
11

SOMEONE PLEASE HELP ME PLEASE!!!

Mathematics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:

193%

Step-by-step explanation:

40.2 - 13.7 = 26.5

26.5 ÷ 13.7 = 1.93

1.93 = 193%

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How much being would you need to be on a skateboard to go on a loop D Loop on and on a ramp
dolphi86 [110]

Answer:

If you know the radius of a circular track, you can use physics to calculate how fast an object needs to move in order to stay in contact with the track without falling when it reaches the top of the loop.

4 0
3 years ago
Read 2 more answers
Need help
aleksandr82 [10.1K]

Using the normal distribution, the probabilities are given as follows:

a. 0.4602 = 46.02%.

b. 0.281 = 28.1%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

The parameters are given as follows:

\mu = 959, \sigma = 263, n = 37, s = \frac{263}{\sqrt{37}} = 43.24

Item a:

The probability is <u>one subtracted by the p-value of Z when X = 984</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{984 - 959}{263}

Z = 0.1

Z = 0.1 has a p-value of 0.5398.

1 - 0.5398 = 0.4602.

Item b:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem:

Z = \frac{X - \mu}{s}

Z = \frac{984 - 959}{43.24}

Z = 0.58

Z = 0.58 has a p-value of 0.7190.

1 - 0.719 = 0.281.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0
1 year ago
Add 5.6 + (-3.2) + (-2.3) PLEASE HELP! This is very time sensitive, have a nice day!
Tpy6a [65]

Answer:

0.1

Step-by-step explanation:

Take paranthesis away.

5.6-3.2-2.3

subtract 3.2 from 5.6

2.4-2.3=0.1

Hope it helps1

3 0
2 years ago
Read 2 more answers
We have 4different boxesand 6different objects. We want to distribute the objects into the boxes such that at no box is empty. I
Musya8 [376]

Answer:

Following are the solution to this question:

Step-by-step explanation:

They provide various boxes or various objects.  It also wants objects to be distributed into containers, so no container is empty.  All we select k objects of r to keep no boxes empty, which (r C k) could be done.  All such k artifacts can be placed in k containers, each of them in k! Forms. There will be remaining (r-k) objects. All can be put in any of k boxes.  Therefore, these (r-k) objects could in the k^{(r-k)} manner are organized.  Consequently, both possible ways to do this are

=\binom{r}{k} \times k! \times k^{r-k}\\\\=\frac{r! \times k^{r-k}}{(r-k)!}

Consequently, the number of ways that r objects in k different boxes can be arranged to make no book empty is every possible one

= \frac{r!k^{r-k}}{(r-k)!}

5 0
3 years ago
Jordan makes two batches of cookies each week for her book club. Each batch of cookies needs 1 2/3 cups of flour. If she starts
mel-nik [20]

Answer:

സ്സ്‌സെക്സെക്സ് sexse

3 0
2 years ago
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